Chapter 3, Problem 3.54AP

An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 800 m, horizontal distance 19.2 km, and 25.0° south of west. The second aircraft is at altitude 1 100 m, horizontal distance 17.6 km, and 20.0° south of west. What is the distance between the two aircraft? (Place the x axis west, the y axis south, and the z axis vertical.)

To determine

The distance between the two aircraft.

Answer to Problem 3.54AP

The distance between the two aircraft is $2.29\text{km}$.

Explanation of Solution

Given info: The air traffic controller observes two aircraft on his radar screen. The first is at altitude $800\text{m}$, horizontal distance is $19.2\text{km}$ and $25.0°$ south of west direction. The second aircraft is at altitude $1100\text{m}$, horizontal distance is $17.6\text{km}$ and $20.0°$ south of west direction.

Consider the $x$ axis as the west direction, $y$ axis as south direction and $z$ axis as vertical direction. The vector component for the aircraft is given below.

Figure (1)

The displacement vector of the first aircraft in the $x$ direction is,

${\overrightarrow{\text{r}}}_{{\text{1}}_{\text{x}}}=r_{\text{1}}\left(\text{cos}{\theta }_1\right)\widehat{\text{i}}$

Here,

${\overrightarrow{\text{r}}}_{1_{\text{x}}}$ is the displacement vector of the first aircraft in $x$ direction.

$r_{\text{1}}$ is the displacement magnitude of the first aircraft.

$\widehat{\text{i}}$ is the unit vector component in x direction.

${\theta }_1$ is the angle of the first aircraft.

Substitute $19.2\text{km}$ for $r_{\text{1}}$ and $25.0°$ for ${\theta }_1$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{r}}}_{\text{1}}=19.2\text{km}\left(\text{cos}\left(25.0°\right)\right)\widehat{\text{i}}\\ =\left(\text{17}\text{.401}\text{km}\right)\widehat{\text{i}}\\ \approx \left(\text{17}\text{.4}\text{km}\right)\widehat{\text{i}}\end{array}$

Thus, the displacement vector of the first aircraft in $x$ direction is $\left(\text{17}\text{.4}\text{km}\right)\widehat{\text{i}}$.

The displacement vector of the first aircraft in the $y$ direction is,

${\overrightarrow{\text{r}}}_{{\text{1}}_{\text{y}}}=r_{\text{1}}\left(\text{sin}{\theta }_1\right)\widehat{\text{j}}$

Here,

${\overrightarrow{\text{r}}}_{{\text{1}}_{\text{y}}}$ is the displacement vector of the first aircraft in $y$ direction.

$\widehat{\text{j}}$ is the unit vector component in $y$ direction.

Substitute $19.2\text{km}$ for $r_{\text{1}}$ and $25.0°$ for ${\theta }_1$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{r}}}_{{\text{1}}_{\text{y}}}=19.2\text{km}\left(\text{sin}\left(25.0°\right)\right)\widehat{\text{j}}\\ =\left(8.11\text{km}\right)\widehat{\text{j}}\end{array}$

Thus, the displacement vector of the first aircraft in $y$ direction is $\left(8.11\text{km}\right)\widehat{\text{j}}$.

The displacement vector of the first aircraft in the $z$ direction is,

${\overrightarrow{\text{r}}}_{{\text{1}}_{\text{z}}}=r_{\text{a}}\widehat{\text{k}}$

Here,

${\overrightarrow{\text{r}}}_{{\text{1}}_{\text{z}}}$ is the displacement vector of the first aircraft in $z$ direction.

$r_{\text{a}}$ is the altitude of the first aircraft.

$\widehat{\text{k}}$ is the unit vector component in $z$ direction.

Substitute $800\text{m}$ for $r_{\text{a}}$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{r}}}_{{\text{1}}_{\text{z}}}=\left(800\text{m}\right)\widehat{\text{k}}\\ =\left(800\text{m}\times \frac{1\text{km}}{1000\text{m}}\right)\widehat{\text{k}}\\ =\left(\text{0}\text{.8}\text{km}\right)\widehat{\text{k}}\end{array}$

Thus, the displacement vector of the first aircraft in $z$ direction is $\left(0.8\text{km}\right)\widehat{\text{k}}$.

The position vector of the first aircraft is,

${\overrightarrow{\text{r}}}_1={\overrightarrow{\text{r}}}_{1_{\text{x}}}+{\overrightarrow{\text{r}}}_{1_{\text{y}}}+{\overrightarrow{\text{r}}}_{1_{\text{z}}}$

Here,

${\overrightarrow{\text{r}}}_1$ is the position vector of the first aircraft.

Substitute $\left(\text{17}\text{.4}\text{km}\right)\widehat{\text{i}}$ for ${\overrightarrow{\text{r}}}_{1_{\text{x}}}$, $\left(8.11\text{km}\right)\widehat{\text{j}}$ for ${\overrightarrow{\text{r}}}_{{\text{1}}_{\text{y}}}$ and $\left(0.8\text{km}\right)\widehat{\text{k}}$ for ${\overrightarrow{\text{r}}}_{{\text{1}}_{\text{z}}}$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{r}}}_1=\left(\text{17}\text{.4}\text{km}\right)\widehat{\text{i}}+\left(8.11\text{km}\right)\widehat{\text{j}}+\left(0.8\text{km}\right)\widehat{\text{k}}\\ =\left[\left(\text{17}\text{.4}\right)\widehat{\text{i}}+\left(8.11\right)\widehat{\text{j}}+\left(0.8\right)\widehat{\text{k}}\right]\text{km}\end{array}$

Thus, the position vector of the first aircraft is $\left[\left(\text{17}\text{.4}\right)\widehat{\text{i}}+\left(8.11\right)\widehat{\text{j}}+\left(0.8\right)\widehat{\text{k}}\right]\text{km}$.

The displacement vector of the second aircraft in the $x$ direction is,

${\overrightarrow{\text{r}}}_{{\text{2}}_{\text{x}}}=r_{\text{2}}\left(\text{cos}{\theta }_2\right)\widehat{\text{i}}$

Here,

${\overrightarrow{\text{r}}}_{{\text{2}}_{\text{x}}}$ is the horizontal displacement vector of the second aircraft in $x$ direction.

$r_2$ is the displacement magnitude of the second aircraft.

${\theta }_2$ is the angle of the second aircraft.

Substitute $17.6\text{km}$ for $r_2$ and $20.0°$ for ${\theta }_2$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{r}}}_{\text{2}}=17.6\text{km}\left(\text{cos}\left(20.0°\right)\right)\widehat{\text{i}}\\ =\left(16.53\text{km}\right)\widehat{\text{i}}\\ \approx \left(16.5\text{km}\right)\widehat{\text{i}}\end{array}$

Thus, the displacement vector of the second aircraft in $x$ direction is $\left(16.5\text{km}\right)\widehat{\text{i}}$.

The displacement vector of the second aircraft in the $y$ direction is,

${\overrightarrow{\text{r}}}_{{\text{2}}_{\text{y}}}=r_{\text{2}}\left(\sin {\theta }_2\right)\widehat{\text{j}}$

Here,

${\overrightarrow{\text{r}}}_{{\text{2}}_{\text{y}}}$ is the horizontal displacement vector of the second aircraft in $y$ direction.

Substitute $17.6\text{km}$ for $r_2$ and $20.0°$ for ${\theta }_2$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{r}}}_{{\text{2}}_{\text{y}}}=17.6\text{km}\left(\sin \left(20.0°\right)\right)\widehat{\text{j}}\\ =\left(6.019\text{km}\right)\widehat{\text{j}}\\ \approx \left(6.02\text{km}\right)\widehat{\text{j}}\end{array}$

Thus, the displacement vector of the second aircraft in $y$ direction is $\left(6.02\text{km}\right)\widehat{\text{j}}$.

The displacement vector of the second aircraft in the $z$ direction is,

${\overrightarrow{\text{r}}}_{2_{\text{z}}}=r_{\text{a}}\text{'}\widehat{\text{k}}$

Here,

${\overrightarrow{\text{r}}}_{2_{\text{z}}}$ is the displacement vector of the second aircraft in $z$ direction.

$r_{\text{a}}\text{'}$ is the altitude of the second aircraft.

Substitute $1100\text{m}$ for $r_{\text{a}}\text{'}$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{r}}}_{2_{\text{z}}}=\left(1100\text{m}\right)\widehat{\text{k}}\\ =\left(1100\text{m}\times \frac{1\text{km}}{1000\text{m}}\right)\widehat{\text{k}}\\ =\left(\text{1}\text{.1}\text{km}\right)\widehat{\text{k}}\end{array}$

Thus, the displacement vector of the second aircraft in $z$ direction is $\left(\text{1}\text{.1}\text{km}\right)\widehat{\text{k}}$.

The position vector of the second aircraft is,

${\overrightarrow{\text{r}}}_2={\overrightarrow{\text{r}}}_{2_{\text{x}}}+{\overrightarrow{\text{r}}}_{2_{\text{y}}}+{\overrightarrow{\text{r}}}_{2_{\text{z}}}$

Here,

${\overrightarrow{\text{r}}}_2$ is the position vector of the second aircraft.

Substitute $\left(16.5\text{km}\right)\widehat{\text{i}}$ for ${\overrightarrow{\text{r}}}_{2_{\text{x}}}$, $\left(6.02\text{km}\right)\widehat{\text{j}}$ for ${\overrightarrow{\text{r}}}_{2_{\text{y}}}$ and $\left(\text{1}\text{.1}\text{km}\right)\widehat{\text{k}}$ for ${\overrightarrow{\text{r}}}_{2_{\text{z}}}$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{r}}}_2=\left(16.5\text{km}\right)\widehat{\text{i}}+\left(6.02\text{km}\right)\widehat{\text{j}}+\left(\text{1}\text{.1}\text{km}\right)\widehat{\text{k}}\\ =\left[\left(16.5\right)\widehat{\text{i}}+\left(6.02\right)\widehat{\text{j}}+\left(\text{1}\text{.1}\right)\widehat{\text{k}}\right]\text{km}\end{array}$

Thus, the position vector of the second aircraft is $\left[\left(16.5\right)\widehat{\text{i}}+\left(6.02\right)\widehat{\text{j}}+\left(\text{1}\text{.1}\right)\widehat{\text{k}}\right]\text{km}$.

The net distance vector between the two aircraft is,

$\overrightarrow{\text{r}}={\overrightarrow{\text{r}}}_2-{\overrightarrow{\text{r}}}_1$

Here,

$\overrightarrow{\text{r}}$ is the net distance vector between the two aircraft.

Substitute $\left[\left(16.5\right)\widehat{\text{i}}+\left(6.02\right)\widehat{\text{j}}+\left(\text{1}\text{.1}\right)\widehat{\text{k}}\right]\text{km}$ for ${\overrightarrow{\text{r}}}_2$ and $\left[\left(\text{17}\text{.4}\right)\widehat{\text{i}}+\left(8.11\right)\widehat{\text{j}}+\left(0.8\right)\widehat{\text{k}}\right]\text{km}$ for ${\overrightarrow{\text{r}}}_1$ in the above equation.

$\begin{array}{c}\overrightarrow{\text{r}}=\left[\left(16.5\right)\widehat{\text{i}}+\left(6.02\right)\widehat{\text{j}}+\left(\text{1}\text{.1}\right)\widehat{\text{k}}\right]\text{km}-\left[\left(\text{17}\text{.4}\right)\widehat{\text{i}}+\left(8.11\right)\widehat{\text{j}}+\left(0.8\right)\widehat{\text{k}}\right]\text{km}\\ =\left[\left(16.5\text{km}\right)-\left(\text{17}\text{.4}\text{km}\right)\right]\widehat{\text{i}}+\left[\left(6.02\text{km}\right)-\left(8.11\text{km}\right)\right]\widehat{\text{j}}+\left[\left(\text{1}\text{.1}\text{km}\right)-\left(0.8\text{km}\right)\right]\widehat{\text{k}}\\ =\left(-0.9\text{km}\right)\widehat{\text{i}}+\left(-2.09\text{km}\right)\widehat{\text{j}}+\left(0.3\text{km}\right)\widehat{\text{k}}\end{array}$

Thus, the net distance vector between the two aircraft is $\left(-0.9\text{km}\right)\widehat{\text{i}}+\left(-2.09\text{km}\right)\widehat{\text{j}}+\left(0.3\text{km}\right)\widehat{\text{k}}$.

The magnitude for the net distance vector between the two aircraft is,

$r=\sqrt{r_{\text{x}}{}^2+r_{\text{y}}{}^2+r_{\text{z}}{}^2}$

Here,

$r_{\text{x}}$ is the net distance vector between the two aircraft in $x$ direction.

$r_{\text{y}}$ is the net distance vector between the two aircraft in $y$ direction.

$r_{\text{z}}$ is the net distance vector between the two aircraft in $z$ direction.

Substitute $-0.9\text{km}$ for $r_{\text{x}}$, $-2.09\text{km}$ for $r_{\text{y}}$ and $0.3\text{km}$ for $r_{\text{z}}$ in the above equation.

$\begin{array}{c}r=\sqrt{{\left(-0.9\text{km}\right)}^2+{\left(-2.09\text{km}\right)}^2+{\left(0.3\text{km}\right)}^2}\\ =2.295\text{km}\\ \approx 2.29\text{km}\end{array}$

Conclusion:

Therefore, the distance between the two aircraft is $2.29\text{km}$.